Q: A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and \displaystyle \angle ADC=130^{\circ} Find \displaystyle \angle ABC (in degrees)

\Rightarrow Quadrilateral ABCD inscribe in a circle. \therefore Quadrilateral ABCD is a cyclic quadrilateral. \Rightarrow \angle ADC+\angle ABC=180^o [ Sum of opposite angles of cyclic quadrilateral is supplementary ] \Rightarrow 130^o+\angle ABC=180^o \Rightarrow \angle ABC=180^o-130^o \Rightarrow \angle ABC=50^o

Question 1

Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle .

Accepted Answer

Given :An arc PQ of a circle subtending angles POQ at the centre O and PAQ at a point A on the remaining part of the circle.To prove :  \angle POQ =2\angle PAQ To prove this theorem we consider the arc AB in three different situations, minor arc AB, major arc AB and semi-circle AB.Construction :Join the line AO extended to B.Proof : \angle BOQ = \angle OAQ+\angle AQO        .....(1)Also, in  	riangle  OAQ, OA=OQ                   [Radii of a circle]Therefore, \angle OAQ = \angle OQA       [Angles opposite to equal sides are equal] \angle BOQ = 2\angle OAQ               .......(2)Similarly,  BOP=2\angle OAP              ........(3)Adding 2 & 3, we get, \angle BOP +\angle BOQ = 2(\angle OAP+\angle OAQ) \angle POQ = 2\angle PAQ                .......(4)For the case 3, where PQ is the major arc, equation 4 is replaced byReflex angle,  \angle POQ=2\angle PAQ

Question 2

PQ  is a chord of length  8  cm of a circle of radius  5  cm. The tangents  P  and  Q  intersect at a point  T . Find the length  TP .

Accepted Answer

Joint  OT . Let it meet  PQ  at the point  R . Then  \Delta TPQ  is isosceles and  TO  is the angle bisector of  \angle PTO .[ \because TP = TQ =  Tangents from  T  upon the circle] 	herefore OT \perp PQ 	herefore OT  bisects  PQ . PR =RQ = 4  cmNow,  OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} = 3  cmNow,  \angle TPR + \angle RPO = 90^{\circ} (\because TPO = 90^{\circ}) = \angle TPR + \angle PTR    (\because TRP = 90^{\circ}) 	herefore \angle RPO = \angle PTR 	herefore  Right triangle  TRP  is similar to the right triangle  PRO . [By A-A Rule of similar triangles] \displaystyle 	herefore \frac{TP}{PO} = \frac{RP}{RO} \Rightarrow \frac{TP}{5} = \frac{4}{3} \displaystyle \Rightarrow TP = \frac{20}{3}  cm.

Question 3

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Accepted Answer

Given: ABCD  is a trapezium where non-parallel sides  AD  and  BC  are equal.
Construction: DM  and  CN  are perpendicular drawn on  AB  from  D  and  C , respectively.
To prove: ABCD  is cyclic trapezium.Proof:
In  	riangle DAM  and  	riangle CBN , AD = BC        ...[Given] ∠AMD = ∠BNC       ...[Right angles] DM = CN      ...[Distance between the parallel lines]Therefore,  	riangle DAM ≅ 	riangle CBN   by  RHS  congruence condition.
Now,  ∠A = ∠B       ...[by CPCT]Also,  ∠B + ∠C = 180°     ....[Sum of the co-interior angles] ⇒ ∠A + ∠C = 180° .
Thus,  ABCD  is a cyclic quadrilateral as the sum of the pair of opposite angles is  180° .

Question 4

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Accepted Answer

Consider the figure.Given, AB  is equal to the radius of the circle.
In  	riangle OAB , OA = OB = AB =  radius of the circle.
Thus,  	riangle OAB  is an equilateral triangle.and  ∠AOC = 60° .Also,  ∠ACB = \dfrac {1}{2}∠AOB = \dfrac {1}{2} × 60° = 30° .Since,  ACBD  is a cyclic quadrilateral, ∠ACB + ∠ADB = 180°  ....[Opposite angles of cyclic quadrilateral are supplementary] ⇒ ∠ADB = 180° - 30° = 150° .
Thus, angle subtend by the chord at a point on the minor arc and also at a point on the major arc are  150°  and  30° , respectively.

Question 5

The perpendicular from the centre of a circle to a chord bisects the chord.

Accepted Answer

To prove that the perpendicular from the centre to a chord bisect the chord.Consider a circle with centre at  O  and  AB  is a chord such that  OX  perpendicular to  AB To prove that    AX = BX In   \Delta OAX   and   \Delta OBX   \angle OXA = \angle OXB    [both are 90 ]   OA = OB    (Both  are radius of circle )   OX = OX    (common side ) ΔOAX\cong ΔOBX AX = BX   (by property of congruent triangles )hence proved.

Question 6

In given fig. O is centre of the circle if  \angle AOC={ 130 }^{ o }  then find  \angle ABC .

Accepted Answer

\angle ADC = \dfrac{1}{2} \angle AOC (Angle subtended by a chord on mayor area is half of that subtended on center) 	herefore \angle ADC = \dfrac{1}{2} 	imes 130 = 65^{\circ} Now, ADCB is cyclic quadrilateral  \Rightarrow   Sum of opposite Angles is  180^{\circ} 	herefore \angle ABC + \angle ADC = 180^{\circ} \angle ABC + 65= 180^{\circ} \angle ABC = 115^{\circ}

Question 7

In Fig 10.4 if  \displaystyle \angle ABC = 20^{\circ}  then  \displaystyle \angle AOC   is equal to

Accepted Answer

We know that the angle at the centre of the circle is twice the angle at the circumference subtended by the same arc.Thus,    \angle AOC=2 	imes \angle ABC \implies \angle AOC=2	imes 20^0 \implies \angle AOC=40^0

Question 8

A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and  \displaystyle \angle ADC=130^{\circ}  Find  \displaystyle \angle ABC   (in degrees)

Accepted Answer

\Rightarrow   Quadrilateral  ABCD  inscribe in a circle. 	herefore   Quadrilateral  ABCD  is a cyclic quadrilateral. \Rightarrow    \angle ADC+\angle ABC=180^o            [ Sum of opposite angles of cyclic quadrilateral is supplementary ] \Rightarrow    130^o+\angle ABC=180^o \Rightarrow    \angle ABC=180^o-130^o \Rightarrow    \angle ABC=50^o

Question 9

ABCD  is a cyclic quadrilateral such that   A={ 90 }^{ o }, B={ 70 }^{ o }, C={ 95 }^{ o }   and  D={ 105 }^{ o }. The statement is:

Accepted Answer

Given,  ABCD  is a cyclic quadrilateral where   \angle A={ 90 }^{ o } , \angle B={ 70 }^{ o }, \angle C={ 95 }^{ o }  &   \angle D={ 105 }^{ o }. Since  ABCD  is a cyclic quadrilateral,then  \angle A+\angle C ={ 180 }^{ o }  and   \angle B+\angle D={ 180 }^{ o } ...... (i) (The sum of the opposite angles of a cyclic quadrilateral is  { 180 }^{ o }) But here,   \angle A+\angle C={ 90 }^{ o }+95^{ o }=185^{ o }  and  \angle B+\angle D={ 70 }^{ o }+{ 10 }5^{ o }=175^{ o }. This is a contradiction to  (i) .That is, the statement is false.Hence, option  B  is correct.

Question 10

Prove that a cyclic parallelogram is a rectangle.

Accepted Answer

Given, ABCD  is a cyclic parallelogram.To prove, ABCD  is a rectangle.Proof: ∠1 + ∠2 = 180°       ...Opposite angles of a cyclic parallelogramAlso, Opposite angles of a cyclic parallelogram are equal.Thus, ∠1 = ∠2 ⇒ ∠1 + ∠1 = 180° ⇒ ∠1 = 90° One of the interior angle of the parallelogram is right angled. Thus,  ABCD  is a rectangle.